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This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.

The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) \subset F_1(A) \subset ... \subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) \cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $gr(A) = \oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.

Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.

His first claim is that $gr(A/I) = \oplus_{i \in \mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) \cap I)$. Following this, he claims that $gr(SL(2)) \cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).

I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!

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We have $$\begin{align*} \text{gr}(A/I) &=& \bigoplus F_i(A/I)/F_{i-1}(A/I) \\ &\cong& \bigoplus (F_i(A)/F_i(A) \cap I)/(F_{i-1}(A)/F_{i-1}(A) \cap I) \\ &\cong& \bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) \cap I) \end{align*}$$

which follows from the fact that $F_{i-1}(A) \cap I \subseteq F_i(A) \cap I$ and I suppose the third isomorphism theorem.

With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) \cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f \in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?

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Yup. Thank you very much! –  Elden Elmanto Mar 21 '12 at 23:08

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